Question

$EFGH$ is a rhombus. Show that $EG$ bisects $\angle E$ and $\angle G$ both.

Answer 1

Here, $EFGH$ is rhombus and $EG$ is diagonal.

In $△EFG,$

$\Rightarrow$ $EF=FG$ [ Sides of rhombus are equal ]

$\Rightarrow$ $\angle 2=\angle 4$ [ Angles opposite to equal sides are equal ] ----- ( 1 )

Now, $EH\parallel FG$ and $EG$ is transversal. [ Opposite sides of rhombus are parallel ]

$\Rightarrow$ $\angle 1=\angle 4$ [ Alternate angles ] ----- ( 2 )

From ( 1 ) and ( 2 ),

$\Rightarrow$ $\angle 1=\angle 2$

$\Rightarrow$ $EG$ bisects $\angle E$

Now, $EF\parallel GH$ and $EG$ is transversal.

$\Rightarrow$ $\angle 2=\angle 3$ [ Alternate angles ] ----- ( 3 )

From ( 1 ) and ( 3 )

$\Rightarrow$ $\angle 4=\angle 3$

$\Rightarrow$ $EG$ bisects $\angle G$

Hence, $EG$ bisects $\angle E$ and $\angle G$